Maths question

  • Thread starter VietDao29
  • Start date
  • #1
VietDao29
Homework Helper
1,423
3
Hi,
I have been trying to solve it for some days,... but I couldnot find out. The equation is:
[tex]a^{2} + b^{2} = c^{2}[/tex]
a, b, c <> 0
I try to call:
[tex]x_{1} = \frac{b}{a}[/tex]
[tex]x_{2} = \frac{c}{a}[/tex]
And I have
[tex]a^2 \times{(1 + x_{1}^{2} - x_{2}^{2})} = 0[/tex]
And I solve for x1 and x2. But I didn't succeed.
Can you suggest me some other ways??
Thanks,
Viet Dao,
 

Answers and Replies

  • #2
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
134
What are you after?
Is it a formula for how pythogorean triplets in general can be found?
Be more specific!
 
  • #3
dextercioby
Science Advisor
Homework Helper
Insights Author
13,070
637
If this must be the case,then:

[tex] a=m^{2}-n^{2} [/tex]

[tex] b=2mn [/tex]

[tex] c=m^{2}+n^{2} [/tex]

Where "m" and "n" are ARBITRARY.

Daniel.
 
  • #4
VietDao29
Homework Helper
1,423
3
arildno said:
What are you after?
Is it a formula for how pythogorean triplets in general can be found?
Be more specific!
I dunno really. My friend just asked me that some weeks ago, and I simply got stuck.
I remember reading a book that says:
[tex]a^{n} + b^{n} = c^{n}[/tex]
Where a, b, c <> 0 and [itex]a, b, c \in Z[/itex]
[itex]n \in N[/itex], n > 2. Then there will be no answer. But it didn't prove it.
--------------
Thanks Dextercioby, I got it now... But how can you know that?
[tex] a=m^{2}-n^{2} [/tex]
[tex] b=2mn [/tex]
[tex] c=m^{2}+n^{2} [/tex]
Is there another case?
--------------
Viet Dao,
 
  • #5
dextercioby
Science Advisor
Homework Helper
Insights Author
13,070
637
I didn't invent it,certainly...I read it somewhere,a long time ago.And it's not too hard to remember.

No,that formula covers all possible "m" and "n" from R (in particular from N hence generating the pythagoreic triplets).

Daniel.
 
  • #6
VietDao29
Homework Helper
1,423
3
Hi,
Thanks a lot dextercioby. :smile: That really helps.
Viet Dao,
 
  • #7
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
I think you need an additional factor to get all triples:

a = p(m2 - n2)
b = 2mnp
c = p(m2 + n2)


Of course, you don't need p for primitive triples.
 
  • #8
dextercioby
Science Advisor
Homework Helper
Insights Author
13,070
637
I see what u mean.Yes,the triplet (6,12,15) would not be found through "m" and "n"...

Daniel.
 
  • #9
VietDao29
Homework Helper
1,423
3
Hi,
Thanks, Hurkyl. I appreciate your help.
Viet Dao,
 
  • #10
dextercioby
Science Advisor
Homework Helper
Insights Author
13,070
637
Just a second.I gave the example with the triplet (6,12,15),which HAS SOLUTIONS,but for "m" and "n" in the reals.So my initial statement is still valid.

Daniel.

P.S.Yes,for all natural number triplets,generated by natural "m" and "n",one has to put the "p".
 

Related Threads on Maths question

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
8
Views
1K
Replies
1
Views
4K
  • Last Post
Replies
2
Views
1K
Replies
7
Views
1K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
916
  • Last Post
Replies
4
Views
1K
Top